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Unformatted text preview: Lecture 10 Linear Systems of First order ODEs Weve looked at x = f ( t,x ), for x R depending on an independent variable t . Recall : If f and f y are continuous, then we can find a solution (through Picards iteration) which is unique if we fix an initial value x at t = 0. For special classes of f , one can say a lot. A particular case of interest is when the ODE is linear , which is the case when f ( t,x ) is linear in x , i.e., f ( t,x ) = a ( t ) x + b ( t ), for suitable functions a ( t ) and b ( t ). It is said to be homogeneous iff the following happens: When x is a solution, any scalar multiple cx is again a solution ; in particular, 0 is a solution. In the linear case we just considered, it is homogeneous exactly when b ( t ) = 0. We say that the linear ODE has constant coecients if a ( t ) and b ( t ) are both independent of t . Hence we have homogeneity and constant coecients iff a ( t ) is independent of t and b ( t ) = 0; in other words, the ODE is of the form x = ax , for some scalar a . Recall : If x = ax , a is constant, the set of all solutions is given by x = { Be at  B any constant } Indeed, x = ax = ( dx dt = 0 x = 0 : equilibrium point ) 1 x dx dt = a, when x = 0 = ( dx/dt ) x dt = a dt = log  x  = at + c =  x  = e at + c x = Be at , with B = e c = 0 But B = 0 is also possible and corresponds to the equilibrium solution x = 0. Hence the claim above, that the general solution of x = ax is x = Be at , where B is any constant. We may think of x as a vector in a space of dimension 1. The set of all solutions is also a 1dimensional vector space, with B as the coordinate. 1 (Brush up on the basics of vector spaces, linear maps, and properties of matrices  including eigenvalues, eigenvectors, and diagonalization  from Ma1b; it will also be good if you know about the Jordan decomposition, which we will discuss later.) Generalization Let t be an independent variable (as before), and let x be a vector in R n . i.e., x = x 1 x 2 . . . x n , whose derivative is another vector in R n : x = d x dt = dx 1 /dt dx 2 /dt . . . dx n /dt We can look at a linear ODE in vector form: x = A ( t ) x (*) where A ( t ) is an n n matrix; A ( t ) = ( a ij ( t )), 1 i,j n . Explicitly ( ) means we have n ODEs dx 1 dt = a 11 ( t ) x 1 + a 12 ( t ) x 2 + + a 1 n ( t ) x n dx 2 dt = a 21 ( t ) x 1 + a 22 ( t ) x 2 + + a 2 n ( t ) x n . . . dx n dt = a n 1 ( t ) x 1 + a n 2 ( t ) x 2 + + a nn ( t ) x n called an n n linear system of ODEs We can say that this system has constant coecients iff A is independent of t , i.e., each a ij is a constant. From now on, assume that we are in the case of constant coecients, and look for solutions x ( t ) of x = A x . It will be of interest to consider the set of all solutions of such a homogeneous linear system of ODEs:...
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 Fall '08
 Makarov,N
 Linear Systems

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